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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals

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Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 10 - Quadrilaterals.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals Page/Excercise MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Correct option: (a)

  

ABCD is a rhombus.

AD BC and AC is the transversal.

DAC = ACB  (alternate angles)

DAC = 50° 

In ΔAOD, by angle sum property,

AOD + DAO + ADO = 180° 

90° + 50° + ADO = 180° 

ADO = 40° 

ADB = 40° 

Solution 5

Solution 6

Solution 7

  

  

  

Solution 8

 

Solution 9

Correct option: (b)

  

DAO + OAB = DAB

DAO + 35° = 90° 

DAO = 55° 

ABCD is a rectangle and diagonals of a rectangle are equal and bisect each other.

OA = OD

ODA = DAO (angles opposte to equal sides are equal)

ODA = 55° 

In DODA, by angle sum property,

ODA + DAO + AOD = 180° 

55° + 55° + AOD = 180° 

AOD = 70° 

Solution 10

Solution 11

Solution 12

  

  

Solution 13

Solution 14

Correct option: (d)

  

In ΔAPB, by angle sum property,

APB + PAB + PBA = 180° 

In ΔCRD, by angle sum property,

CRD + RDC + RCD = 180° 

Now, SPQ + SRQ = APB + CRD

= 360° - 180° 

= 180° 

Now, PSR + PQR = 360° - (SPQ + SRQ)

= 360° - 180° 

= 180° 

Hence, PQRS is a quadrilateral whose opposite angles are supplementary. 

Solution 15

  

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Correct option: (d)

  

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii),

PQ RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM NO and PN MO.

PMON is a parallelogram.

MPN = MON (opposite angles of a parallelogram are equal)

MPN = BOA (since BOA = MON)

MPN = 90° (since AC BD, BOA = 90°)

QPS = 90° 

Thus, PQRS is a parallelogram whose one angle, i.e. QPS = 90°.

Hence, PQRS is a rectangle if AC BD. 

Solution 21

Correct option: (c)

  

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

PQ RS and QR SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus if diagonals of ABCD are equal. 

Solution 22

Correct option: (d)

  

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

 

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

PMON is a parallelogram.

MPN = MON (opposite angles of a parallelogram are equal)

MPN = BOA (since BOA = MON)

MPN = 90° (since AC BD, BOA = 90°)

QPS = 90° 

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and QPS = 90°.

Hence, PQRS is a square if diagonals of ABCD are equal and perpendicular. 

Solution 23

  

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Correct option: (b)

  

In ΔABC, D and E are the mid-points of sides AB and AC respectively.

Hence, DENM is a parallelogram.

Solution 32

  

  

Solution 33

  

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

 

Solution 39

Solution 40

Solution 41

Solution 42

Since opposite angles of quadrilateral are equal, PQRS is a parallelogram.

PQ = SR (opposite sides of parallelogram are equal)

PQ = 2 cm

Solution 43

The given statement is false.

Diagonals of a parallelogram bisect each other. 

Solution 44

In quadrilateral PQRS, P and S are adjacent angles.

Since the sum of adjacent angles 180°, PQRS is not a parallelogram.

Hence, PQRS is a trapezium. 

Solution 45

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are acute, the sum will be less than 360°. 

Solution 46

The given statement is true.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are right angles,

Sum of all angles of a quadrilateral = 4 × 90° = 360° 

Solution 47

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are obtuse, the sum will be more than 360°. 

Solution 48

We know that the sum of all the four angles of a quadrilateral is 360°.

Here,

70° + 115° + 60° + 120° = 365° 360° 

Hence, we cannot form a quadrilateral with given angles. 

Solution 49

A quadrilateral whose all angles are equal is a rectangle. 

Solution 50

D and E are respectively the midpoints of the sides AB and BC of ΔABC.

Thus, by mid-point theorem, we have

  

Solution 51

Since the diagonals PR and QS of quadrilateral PQRS bisect each, PQRS is a parallelogram.

Now, adjacent angles of parallelogram are supplementary.

Q + R = 180° 

56° + R = 180° 

R = 124° 

Solution 52

AFDE is a parallelogram

AF = ED …(i)

BDEF is a parallelogram.

FB = ED …(ii)

From (i) and (ii),

AF = FB 

Solution 53

Correct option: (a)

If the diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a parallelogram.

So, I gives the answer.

If the diagonals are equal, then the quad. ABCD is a parallelogram.

So, II gives the answer.

Solution 54

Correct option: (c)

If the quad. ABCD is a gm, it could be a rectangle or square or rhombus.

So, statement I is not sufficient to answer the question.

If the diagonals AC and BD are perpendicular to each other, then the gm could be a square or rhombus.

So, statement II is not sufficient to answer the question.

However, if the statements are combined, then the quad. ABCD is a rhombus.

Solution 55

Correct option: (c)

If the diagonals of a gm ABCD are equal, then gm ABCD could either be a rectangle or a square.

If the diagonals of the gm ABCD intersect at right angles, then the gm ABCD could be a square or a rhombus.

However, if both the statements are combined, then gm ABCD will be a square.

Solution 56

Correct option: (b)

If the opposite sides of a quad. ABCD are equal, the quadrilateral is a parallelogram.

If the opposite angles are equal, then the quad. ABCD is a parallelogram.

Solution 57

Solution 58

  

    

 

The Reason (R) is true and is the correct explanation for the Assertion (A).

Solution 59

  

 

Solution 60

Solution 61

Solution 62

Solution 63

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals Page/Excercise 10A

Solution 1

Let the measure of the fourth angle = x° 

For a quadrilateral, sum of four angles = 360° 

x° + 75° + 90° + 75° = 360° 

x° = 360° - 240° 

x° = 120° 

Hence, the measure of fourth angle is 120°. 

Solution 2

Solution 3

Since AB || DC

Solution 4

Given:

 

Solution 5


Solution 6


Solution 7

 

Solution 8

Given: O is a point within a quadrilateral ABCD

 

Solution 9

Given: ABCD is a quadrilateral and AC is one of its disgonals.

Solution 10

Given: ABCD is a quadrilateral.

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals Page/Excercise 10B

Solution 1

Solution 2

 

Solution 3

ABCD is a parallelogram.

Hence, AD || BC.

DAM = AMB (alternate angles)

BAM = AMB (since BAM = DAM)

BM = AB (sides opposite to equal angles are equal)

But, AB = CD (opposite sides of a parallelogram)

BM = AB = CD ….(i)

  

Solution 4

 

 

 

 

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

 


Solution 11

 

Solution 12


Solution 13

 


 

 

Solution 14

  

Let the altitude from D to the side AB bisect AB at point P.

Join BD.

In ΔAMD and ΔBMD,

AM = BM (M is the mid-point of AB)

AMD = BMD (Each 90°)

MD = MD (common)

ΔAMD ΔBMD (by SAS congruence criterion)

AD = BD (c.p.c.t.)

But, AD = AB (sides of a rhombus)

AD = AB = BD

ΔADB is an equilateral triangle.

A = 60° 

C = A = 60° (opposite angles are equal)

B = 180° - A = 180° - 60° = 120° 

D = B = 120° 

Hence, in rhombus ABCD, A = 60°, B = 120°, C = 60° and D = 120°.

Solution 15

Solution 16

  

In ΔABC and ΔADC,

AB = AD (sides of a rhombus are equal)

BC = CD (sides of a rhombus are equal)

AC = AC (common)

ΔABC ΔADC (by SSS congruence criterion)

BAC = DAC and BCA = DCA (c.p.c.t.)

AC bisects A as well as C.

Similarly,

In ΔBAD and ΔBCD,

AB = BC (sides of a rhombus are equal)

AD = CD (sides of a rhombus are equal)

BD = BD (common)

ΔBAD ΔBCD (by SSS congruence criterion)

ABD = CBD and ADB = CDB (c.p.c.t.)

BD bisects B as well as D.

Solution 17

In ΔAMO and ΔCNO

MAO = NCO (AB CD, alternate angles)

AM = CN (given)

AOM = CON (vertically opposite angles)

ΔAMO ΔCNO (by ASA congruence criterion)

AO = CO and MO = NO (c.p.c.t.)

AC and MN bisect each other.

Solution 18


Solution 19

 

Solution 20

DCM = DCN + MCN

90° = DCN + 60° 

DCN = 30° 

In ΔDCN,

DNC + DCN + D = 180° 

90° + 30° + D = 180° 

D = 60° 

B = D = 60° (opposite angles of parallelogram are equal)

A = 180° - B = 180° - 60° = 120° 

C = A = 120° 

Thus, the angles of a parallelogram are 60°, 120°, 60° and 120°.

Solution 21

  

(i) ABCD is a rectangle in which diagonal AC bisects A as well as C.

BAC = DAC ….(i)

And BCA = DCA ….(ii)

Since every rectangle is a parallelogram, therefore

AB DC and AC is the transversal.

BAC = DCA (alternate angles)

DAC = DCA [From (i)]

Thus, in ΔADC,

AD = CD (opposite sides of equal angles are equal)

But, AD = BC and CD = AB (ABCD is a rectangle)

AB = BC = CD = AD

Hence, ABCD is a square.

 

(ii) In ΔBAD and ΔBCD,

AB = CD

AD = BC

BD = BD

ΔBAD ΔBCD (by SSS congruence criterion)

ABD = CBD and ADB = CDB (c.p.c.t.)

Hence, diagonal BD bisects B as well as D.

Solution 22

Solution 23

Solution 24

  

l m and t is a transversal.

APR = PRD (alternate angles)

SPR = PRQ (PS and RQ are the bisectors of APR and PRD)

Thus, PR intersects PS and RQ at P and R respectively such that SPR = PRQ i.e., alternate angles are equal.

PS RQ

Similarly, we have SR PQ.

Hence, PQRS is a parallelogram.

Now, BPR + PRD = 180° (interior angles are supplementary)

2QPR + 2QRP = 180° (PQ and RQ are the bisectors of BPR and PRD)

QPR + QRP = 90° 

In ΔPQR, by angle sum property,

PQR + QPR + QRP = 180° 

PQR + 90° = 180° 

PQR = 90° 

Since PQRS is a parallelogram,

PQR = PSR

PSR = 90° 

Now, SPQ + PQR = 180° (adjacent angles in a parallelogram are supplementary)

SPQ + 90° = 180° 

SPQ = 90° 

SRQ = 90° 

Thus, all the interior angles of quadrilateral PQRS are right angles.

Hence, PQRS is a rectangle.

Solution 25

  

AK = BL = CM = DN (given)

BK = CL = DM = AN (i)(since ABCD is a square)

In ΔAKN and ΔBLK,

AK = BL (given)

A = B (Each 90°)

AN = BK [From (i)]

ΔAKN ΔBLK (by SAS congruence criterion)

AKN = BLK and ANK = BKL (c.p.c.t.)

But, AKN + ANK = 90° and BLK + BKL = 90° 

AKN + ANK + BLK + BKL = 90° + 90° 

2AKN + 2BKL = 180° 

AKN + BKL = 90° 

NKL = 90° 

Similarly, we have

KLM = LMN = MNK = 90° 

Hence, KLMN is a square.

Solution 26

Solution 27

 

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 10 - Quadrilaterals Page/Excercise 10C

Solution 1

  

(i) In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

(ii) In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii), we have

PQ = SR and PQ SR

(iii) Thus, in quadrilateral PQRS, one pair of opposite sides are equal and parallel.

Hence, PQRS is a parallelogram. 

Solution 2

  

Let ΔABC be an isosceles right triangle, right-angled at B.

AB = BC

Let PBSR be a square inscribed in ΔABC with common B.

PB = BS = SR = RP

Now, AB - PB = BC - BS

AP = CS ….(i)

In ΔAPR and ΔCSR

AP = CS  [From (i)

APR = CSR (Each 90°)

PR = SR (sides of a square)

ΔAPR ΔCSR (by SAS congruence criterion)

AR = CR (c.p.c.t.)

Thus, point R bisects the hypotenuse AC.

Solution 3


 

Solution 4

  

In ΔAOM and ΔCON

MAO = OCN  (Alternate angles)

AO = OC (Diagonals of a parallelogram bisect each other)

AOM = CON  (Vertically opposite angles)

ΔAOM ΔCON  (by ASA congruence criterion)

MO = NO (c.p.c.t.)

Thus, MN is bisected at point O. 

Solution 5

Construction: Join diagonal QS. Let QS intersect MN at point O.

  

PQ SR and MN PQ

PQ MN SR

By converse of mid-point theorem a line drawn, through the mid-point of any side of a triangle and parallel to another side bisects the third side. 

Now, in ΔSPQ

MO PQ and M is the mid-point of SP

So, this line will intersect QS at point O and O will be the mid-point of QS.

Also, MN SR

Thus, in ΔQRS, ON SR and O is the midpoint of line QS.

So, by using converse of mid-point theorem, N is the mid-point of QR.

Solution 6

  

PM is the bisector of P.

QPM = SPM ….(i)

PQRS is a parallelogram.

PQ SR and PM is the transversal.

QPM = MS (ii)(alternate angles)

From (i) and (ii),

SPM = PMS ….(iii)

MS = PS = 9 cm (sides opposite to equal angles are equal)

Now, RMT = PMS (iv)(vertically opposite angles)

Also, PS QT and PT is the transversal.

RTM = SPM

RTM = RMT

RT = RM (sides opposite to equal angles are equal)

RM = SR - MS = 12 - 9 = 3 cm

RT = 3 cm

Solution 7

Solution 8

Solution 9


Solution 10

 

Solution 11

Solution 12


Solution 13

 


Solution 14

 


 

 

Solution 15

Solution 16

  

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

PQ = QR = RS = SP  [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus.

Solution 17

  

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii),

PQ || RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM || NO and PN || MO.

PMON is a parallelogram.

MPN = MON (opposite angles of a parallelogram are equal)

MPN = BOA (since BOA = MON)

MPN = 90° (since AC BD, BOA = 90°)

QPS = 90° 

Thus, PQRS is a parallelogram whose one angle, i.e. QPS = 90°.

Hence, PQRS is a rectangle. 

Solution 18

  

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

 

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

PMON is a parallelogram.

MPN = MON (opposite angles of a parallelogram are equal)

MPN = BOA (since BOA = MON)

MPN = 90° (since AC BD, BOA = 90°)

QPS = 90° 

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and QPS = 90°.

Hence, PQRS is a square.

TopperLearning provides step-by-step solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 10 - Quadrilaterals for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 10 - Quadrilaterals now.

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